使用“readr”函数“parse\u number”正确提取数字[重复]

library(stringr)
x <- str_extract_all(strLine,"\\(?[0-9,.]+\\)?")[[1]]
> x
[1] "0"       "3,000"   "(500)"   "0"       "2.25"    "(1,200)"

将parens改为否定:

x <- gsub("\\((.+)\\)","-\\1",x)
x
[1] "0"      "3,000"  "-500"   "0"      "2.25"   "-1,200"

as.numeric() 或 taRifx::destring next version 属于 destring 默认情况下支持负片,因此 keep 不需要选项):

library(taRifx)
destring( x, keep="0-9.-")
[1]    0 3000  -500    0    2.25 -1200

或:

as.numeric(gsub(",","",x))
[1]     0  3000  -500     0     2.25 -1200

这是基本的R方法,为了完整性。。。

x <- unlist(regmatches(strLine, gregexpr('\\(?[0-9,.]+', strLine)))
x <- as.numeric(gsub('\\(', '-', gsub(',', '', x)))
[1]     0.00  3000.00  -500.00     0.00     2.25 -1200.00

当我在一个 data frame (同一列中每行一个字符串)如下:

library(taRifx)
DataFrame$Numbers<-as.character(destring(DataFrame$Strings, keep="0-9.-"))

结果在同一列的新列中 数据帧

既然这是另一个问题 stringi 解决方案(与 stringr

as.numeric(
  stringi::stri_replace_first_fixed(
    stringi::stri_replace_all_regex(
      unlist(stringi::stri_match_all_regex(
        "The transactions (on your account) were as follows: 0 3,000 (500) 0 2.25 (1,200)", 
        "\\(?[0-9,.]+\\)?"
      )), "\\)$|,", ""
    ),
    "(", "-"
  )
)